Question

我仅在条件满足时才打开下拉列表。我怎样才能做到这一点？现在我的代码如下：

<div class="btn-group" uib-dropdown>
  <button class="btn btn-secondary" uib-dropdown-toggle ng-click="vm.openDropdown()">
  open
</button>
  <div class="dropdown-menu dropdown-menu-right" uib-dropdown-menu>
    <a class="dropdown-item d-inline-flex" ng-repeat="test in vm.tests">{{test}}</a>
  </div>
</div>

vm.openDropdown = openDropdown
vm.tests = [1, 2, 3, 4]

function openDropdown () {
    if (vm.test == true) {
////and here I need have condition if something is true, I do not want to open dropdown
     }
 }

Answer 1

您是否尝试过使用ng-disabled ：（如果变量test为true，则无法打开下拉菜单）

<div class="btn-group" uib-dropdown>
    <button class="btn btn-secondary" uib-dropdown-toggle ng-disabled="vm.test">open
    </button>
    <div class="dropdown-menu dropdown-menu-right" uib-dropdown-menu>
       <a class="dropdown-item d-inline-flex" ng-repeat="test in vm.tests">{{test}}</a>
    </div>
</div>

在angularjs中如何在条件下手动打开下拉菜单？

1 个答案: